Ben and Bill

Date: 12/21/97 at 10:21:53
From: SHAMALA
Subject: Algebra
Bill + Ben's Age = 91
Bill is twice as old as Ben was when Bill was as old as Ben is now.
I've tried to form an equation but I get stuck in the first phrase.
Please help.
Thank you,
Shamala

Date: 01/06/98 at 13:32:06
From: Doctor Dawn
Subject: Re: Algebra
Dear Shamala,
Wow! This was a tough question and had me stumped for a while. At
first I was trying to assign Ben a variable and Bill a different
variable, but I couldn't figure out what to put down for the second
sentence. Since the difference in their ages is constant, I finally
decided to assign the difference in their ages a variable. I also know
that Bill is older. Here goes:
First, define the variables
x = difference in Bill's and Ben's ages
b = Ben's age
This means
b + x = Bill's age
and
Bill - x = Ben's age. (This will be important later)
Next write the English sentences as math equations using the
variables.
Bill + Ben = 91
(b + x) + b = 91
so 2b + x = 91. (Equation 1)
Now for the second sentence -
Bill is twice as old as Ben was, when Bill was as old as Ben is now.
The first Bill is now in the present and is represented by b + x
The second Bill is in the past when he was Ben's age now which is b.
Ben in the past was x years younger than Bill in the past, so Ben in
the past is represented by:
Bill (past) - x = b - x.
The second equation then goes like this.
Bill = twice as old as Ben was
(b + x) = 2 * (b - x) (Equation 2)
Now take both equations and solve.
2b + x = 91
b + x = 2*(b - x)
Working with equation 2:
b+x = 2b-2x (distribute the 2)
3x = b (add 2x and subtract b on each side)
Substituting 3x for b into equation 1:
2*3x + x = 91
6x + x = 91
7x = 91
x = 13
The difference in their ages is 13.
Using equation 1 again, and knowing x = 13:
2b + 13 = 91
2b = 78
b = 39 (Ben's age)
Bill = b + x = 39 + 13 = 52.
Bill is 52 years old; Ben is 39 years old. To check, when Bill was 39
(Ben's age now), Ben was 26. 52 is twice 26.
-Doctor Dawn, The Math Forum
Check out our web site! http://mathforum.org/dr.math/